FLAP M6.1 Introducing differential equations COPYRIGHT 1998 THE OPEN UNIVERSITY S570 V1.1

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1 F1 (a) This equation is of first order and of first degree. It is a linear equation, since the dependent variable y and its derivative only appear raised to the first power, and there are no products of y and its derivative present. (b) This equation is of first order and of first degree. It is not linear, since it contains the product y(dy/dx). (c) This equation is of second order and of second degree. It is not linear, since it contains higher powers of the derivatives of the dependent variable.

2 F2 (a) To construct the differential equation we use the one-dimensional form of Newton s second law, F x = ma x, taking acceleration a x and total force F x to be positive if they are directed downwards. By definition, the acceleration is equal to the rate of change of velocity v x so that a x = dv x1 /dt. The total force is given by the sum of two terms. The first is the force due to gravity, +mg (it has a + sign since it acts downwards). The second is a resistive force, the direction of which is opposite to that of v x ; it can be written as kv x, where the constant k is positive. Substituting these expressions into F x = ma x gives the required result. (b) A general solution to a linear differential equation of order n is a solution that contains n independent arbitrary constants. Here, we have a first-order linear differential equation, so its general solution should contain one arbitrary constant. The proposed solution does contain one arbitrary constant, A, so it will be a general solution provided that it does actually satisfy the equation, and we must now check this. Differentiating the expression for v x, we find dv x dt = ka m e kt m so that m dv x dt = kae kt m Also, mg kv x = mg k(mg/k + Ae kt/m ) = kae kt/m. So when we substitute the given expression for v x into the equation, we obtain an identity. Hence this expression for v x is a general solution.

3 (c) To determine the value of A, we use the initial condition v x = 0 at t = 0. Substituting these values into the expression given for v x, we find 0 = mg/k + A3or3A = mg/k As t becomes very large, e kt/m tends to zero; so the limiting value of v x is mg/k.

4 Rl We first divide both sides by e x, and since 1/e x = e x, the result can be written as xe x = 1. Now multiply both y + 1 sides by (1y + 1), to give xe x = y + 1. Subtracting 1 from both sides gives the required answer: y = xe x 1 (For further information about rearrangement of an equation and exponential function consult the Glossary.)

5 R2 (a) The values are: e 0 = l; sin10 = 0; cos10 = 1. (b) The values are: sin1(π/2) = 1; cos1(π/2) = 0; sin1π = 0; cos1π = 1. Note that here the argument of the trigonometric functions is expressed in radians, as is virtually always the case in calculus. (For further information about exponential function and trigonometric functions consult the Glossary.)

6 R3 If f(x) = x 3, then f(x + δ1x) = (x + δ1x) 3 = x 3 + 3x 2 δ1x + 3x(δ1x) 2 + (δ1x) 3 So f ( x + δx ) f ( x ) δx = 3x 2 + 3xδ1x + (δ1x) 2 = 3x 2 δx + 3x(δx ) 2 + (δx ) 3 δx Taking the limit δx 0, we find df dx = 3x 2. (For further information about derivatives consult the Glossary.)

7 R4 (a) To calculate the derivative of x1log e 1x, we use the product rule, d [ dx f ( x )g( x )] = df dg g( x ) + f ( x ) dx dx Taking f1(x) = x and g(x) = log e 1x, it follows that df1/dx = 1 and dg/dx = 1/x. d Thus dx ( x log e x ) = 1 log e x + x 1 x = 1 + log e x Notice that we prefer to write 1 + log e 1x rather than log e 1x + 1 since the latter might be confused with log e 1(x + 1). (b) To calculate the derivative of (sin1x)/x, we use the quotient rule, d [ dx f ( x ) g( x ) g(x)(df dx ) f ( x )(dg dx ) ] = g( x ) [ ] 2 taking f1(x) = sin1x, g(x) = x. Then df1/dx = cos1x, dg/dx = 1. Thus we find d [ dx (sin x ) x xcos x sin x ] = x 2

8 (c) To calculate the derivative of d dx f[ y(x) ] = df dy dy dx 1 3x, we use the chain rule, taking f1(1y) = y and y(x) = 1 3x. Then df/dy = 1/(2 y ) = 1/(2 1 3x ) and dy/dx = 3. Thus the derivative is x (If you feel unsure of any of these terms, refer to the Glossary.)

9 R5 Differentiating y 2 implicitly with respect to x gives d dx ( y2 ) = 2 y dy dx = 2 ye x x (For further information about implicit differentiation consult the Glossary.)

10 R6 (a) The first derivative of x 4 is 4x 3 ; differentiating again gives the second derivative, 12x 2. (b) The first derivative of 3e 2x is 6e 2x ; differentiating again gives the second derivative, 12e 2x. (If you feel unsure of these terms, refer to the Glossary.)

11 R7 (a) x n dx = x n+1 (n + 1) + constant so, with n = 1/3 we get (b) [ 1(3x) ] dx = (1 3) (1 x)dx = (1 3) log e x + constant x 1/3 dx = (3 4)x 4/3 + constant (c) To find e nx dx let u = nx, then e nx dx = e u (dx du)du = (1 n) e u du so e nx dx = (1 n)e nx + constant Comment The above integrals are based on standard results. If you had difficulty with them, you should read the modules on integration.

12 T1 We are told the rate of change of temperature with time, dt/dt, is negative and proportional to (T T 0 ). Thus dt = k(t T 0 ) dt where k is a positive constant.

13 T2 The rate of change of the number n of people with a cold is dn/dt. If the population of the community is fixed at N 0, the number of people who do not have a cold is (N 0 n). The derivative dn/dt is positive and proportional to the product of n and (N 0 n), and so we have dn = kn( N 0 n) dt where k is a positive constant.

14 T3 Since the probability that a molecule makes a collision in distance δx is approximately Cδx, the probability that it does not make a collision in this distance is approximately (1 Cδx). We multiply this by N(x) to obtain an approximate value for N(x + δx) so that N(x + δx) N(x)(1 Cδx) If we rearrange this approximate equation we get N ( x + δx ) N ( x ) CN ( x ) δx When we let δx tend to zero, the approximation becomes exact, and the quantity on the left hand-side becomes dn/dx. So the required differential equation is dn dx = CN

15 T4 Equation 4, dn (t ) = λn(t) (Eqn 4) dt is first order; only the first derivative of the dependent variable appears. Similarly Equation 7, dn dt (t ) = kn(t ) N 0 n(t ) [ ] (Eqn 7) is first order. In Equation 16, m d 2 x dt 2 = kx C dx dt (Eqn 16) the highest order derivative appearing is the second derivative of x; therefore it is second order.

16 T5 (a) In this equation, the highest derivative present is d 2 y/dx 2, and it is raised to the first power. The equation is therefore of first degree. (b) In this equation, the highest derivative present is dy/dx, and it is raised to the power 2. The equation is therefore of second degree. Note that in this case there is no need to rationalize the power of y appearing in the equation. It is only when derivatives of the dependent variable appear raised to fractional powers that the equation needs to be rationalized.

17 T6 Equations 4 and 16 are linear. dn (t ) = λn(t) (Eqn 4) dt m d 2 x dt 2 = kx C dx dt (Eqn 16) The dependent variable and its derivative appear raised only to the first power, and there are no products or functions of the dependent variable or its derivatives with one another in the equations. Equation 7 is not linear; dn dt (t ) = kn(t ) N 0 n(t ) [ ] (Eqn 7) there is a term containing the square of the dependent variable on the right-hand side.

18 Equation 13 is not linear; dy = a 2gy (Eqn 13) dt A the square root of the dependent variable appears on the right-hand side. Equation 17 is not linear, y dy = x (Eqn 17) dx due to the presence of the product of the dependent variable and its first derivative.

19 T7 If y = e x, then dy/dx = 1e x. Thus substituting y and its derivative into the differential equation gives us 1e x = 1e x, which is an identity. So y = e x is a solution.

20 T8 If y = Ce x, then dy/dx = Ce x. Thus substituting y and its derivative into the differential equation gives us Ce x = Ce x, which is an identity. So y = Ce x is a general solution.

21 T9 To see whether y = Ae (x1+1b) is a solution, we substitute it into the equation. Note that we can write this function as Ae B e x, and that Ae B is simply a constant; so we have dy dx = Ae B e x and d 2 y dx 2 = Ae B e x On substitution, we find that the left-hand side of the equation in Question T9 becomes Ae B e x 3Ae B e x + 2Ae B e x which is identically equal to zero. Thus we have a solution. But we have seen that we can write this solution as y = Ae B e x, where Ae B is a constant that we may call C. Thus the two constants appearing here are not independent; the solution can be written as y = Ce x, showing explicitly that it contains only one essential constant. It therefore cannot be the general solution to this second-order equation, which must contain two essential constants.

22 T10 We first substitute the values t = 0, x = 0.011m into the expression for x, which gives 0.011m = A1cos10 + B1sin10, i.e. A = 0.011m. Differentiating x to find the velocity as a function of time t gives us dx/dt = A1sin1t + B1cos1t, and substituting the values t = 0, dx/dt = 0 gives us 0 = A1sin10 + B1cos10, i.e. B = 0. Thus the values of the constants are A = 0.011m, B = 0.

23 T11 With g = l01m1s 2 Equation 33 x = gt u xt + x 0 (Eqn 33) gives x = (51m1s 2 )t 12 + (151m1s 1 )t + 11m and v x = dx/dt = (101m1s 2 )t + 151m1s 1 Substituting t = 21s into these equations, we find x = 111m, v x = 51m1s 1. When the ball returns to its starting point, x = 11m. Thus (51m1s 2 )t 12 + (151m1s 1 )t = 0. This equation is satisfied by t = 0 or by t = 31s. So the ball returns to its starting point at t = 31s. Its velocity at this point is 151m1s 1.

24 T12 With g = l01m1s 2, we have x = (5.0001m1s 2 )t 12 + ( m1s 1 )t m and v x = dx/dt = (l0.0001m1s 1 )t m1s 1 Substituting t = 21s into these equations, we obtain x = m, v x = m1s 1. These differ from the results obtained in Question T12 by 0.3%. When the ball returns to its starting point, x = m. Thus (5.0001m1s 2 )t 12 + ( m1s 1 )t = 0. This equation is satisfied by t = 0 or by t = s. So the ball returns to its starting point at t = s, which differs from the answer obtained in Question T11 by 0.1%.